Two cars are driving towards an intersection from perpendicular directions. The first car's velocity is $10$ meters per second and the second car's velocity is $6$ meters per second. At a certain instant, the first car is $4$ meters from the intersection and the second car is $3$ meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)? Choose 1 answer: Choose 1 answer: (Choice A) A $-\sqrt{136}$ (Choice B) B $-11.6$ (Choice C) C $-10.8$ (Choice D) D $-5$
Solution: Setting up the math Let... $a(t)$ denote the distance between the first car and the intersection at time $t$, $b(t)$ denote the distance between the second car and the intersection at time $t$, and $c(t)$ denote the distance between the two cars at time $t$. $a(t)$ $b(t)$ $c(t)$ We are given that $a'(t)=-10$ and $b'(t)=-6$. Notice that both $a'$ and $b'$ are negative since the cars are getting closer to the intersection. We are also given that $a(t_0)=4$ and $b(t_0)=3$ for a specific time $t_0$. We want to find $c'(t_0)$. Relating the measures The measures relate to each other through the Pythagorean theorem: $[a(t)]^2+[b(t)]^2=[c(t)]^2$ We can differentiate both sides to find an expression for $c'(t)$ : $c'(t)=\dfrac{1}{c(t)}\left[a(t)a'(t)+b(t)b'(t)\right]$ Using the information to solve In order to find $c'(t_0)$ we need to find $c(t_0)$. Using the Pythagorean theorem and the fact that $a(t_0)=4$ and $b(t_0)=3$, we can find that $c(t_0)=5$. Let's plug ${c(t_0)}={5}$, ${a(t_0)}={4}$, ${a'(t_0)}={-10}$, $C{b(t_0)}=C{3}$, and ${b'(t_0)}={-6}$ into the expression for $c'(t_0)$ : $\begin{aligned} c'(t_0)&=\dfrac{1}{{c(t_0)}}\left[{a(t_0)}{a'(t_0)}+C{b(t_0)}{b'(t_0)}\right] \\\\ &=\dfrac{1}{({5})}\left[({4})({-10})+(C{3})({-6})\right] \\\\ &=-11.6 \end{aligned}$ In conclusion, the rate of change of the distance between the two cars at that instant is $-11.6$ meters per second. Since the rate of change is negative, we know that the distance is decreasing.